Calculate the Area: Finding Ascending Region of f(x)=6x²-12

Function Derivatives with Increasing Intervals

Find the ascending area of the function

f(x)=6x212 f(x)=6x^2-12

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the domain of increase of the function
00:04 Let's examine the trinomial coefficients
00:10 Let's examine the coefficient of X squared, it's positive so the function is happy
00:16 We'll use the formula to find the vertex of the parabola
00:19 We'll substitute appropriate values according to the given data and solve to find the vertex
00:22 This is the X value at the vertex of the parabola
00:30 We'll determine when the parabola is decreasing and when it's increasing based on its type
00:38 We'll draw the X-axis and find the domain of increase
00:41 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the ascending area of the function

f(x)=6x212 f(x)=6x^2-12

2

Step-by-step solution

To determine the ascending area of the function f(x)=6x212 f(x) = 6x^2 - 12 , we will follow these steps:

  • Step 1: Calculate the derivative of the given function.
  • Step 2: Set the inequality f(x)>0 f'(x) > 0 to find the interval where the function is increasing.
  • Step 3: Solve the inequality for x x .

Let's begin with Step 1:
The derivative of f(x)=6x212 f(x) = 6x^2 - 12 with respect to x x is:

f(x)=ddx(6x212)=12x f'(x) = \frac{d}{dx}(6x^2 - 12) = 12x .

Step 2: We need to find where 12x>0 12x > 0 . This requires:

x>0 x > 0 .

Step 3: Therefore, the function f(x)=6x212 f(x) = 6x^2 - 12 is increasing when x>0 x > 0 .

Thus, the increasing interval of the function is when x>0 x > 0 .

The solution to the problem is 0<x 0 < x .

3

Final Answer

0<x 0 < x

Key Points to Remember

Essential concepts to master this topic
  • Rule: Function increases when derivative f'(x) > 0
  • Technique: Find derivative: f'(x) = 12x, then solve 12x > 0
  • Check: Test point x = 1: f'(1) = 12 > 0, so increasing ✓

Common Mistakes

Avoid these frequent errors
  • Confusing increasing with decreasing regions
    Don't solve f'(x) < 0 when looking for increasing intervals = gives decreasing region instead! This reverses the answer completely. Always solve f'(x) > 0 for increasing intervals and f'(x) < 0 for decreasing intervals.

Practice Quiz

Test your knowledge with interactive questions

Find the ascending area of the function

\( f(x)=2x^2 \)

FAQ

Everything you need to know about this question

Why do I need to find the derivative first?

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The derivative tells you the slope of the function at any point! When f(x)>0 f'(x) > 0 , the slope is positive, meaning the function is going upward (increasing).

What does 'ascending area' or 'increasing interval' mean?

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It's the range of x-values where the function goes up as you move from left to right. Think of it like climbing a hill - you're ascending when moving upward!

How do I know if x > 0 or x < 0 is correct?

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After finding f(x)=12x f'(x) = 12x , solve 12x>0 12x > 0 . Since 12 is positive, divide both sides by 12 to get x>0 x > 0 .

Can I test a specific point to verify my answer?

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Yes! Pick any point in your interval. For example, if x>0 x > 0 , test x=1 x = 1 : f(1)=12(1)=12>0 f'(1) = 12(1) = 12 > 0

What if the derivative equals zero?

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When f(x)=0 f'(x) = 0 , the function has a horizontal tangent - it's neither increasing nor decreasing at that point. Here, f(0)=0 f'(0) = 0 at x=0 x = 0 .

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